2013 USAJMO Problems/Problem 1

Problem

Are there integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?

Solution

No, such integers do not exist. This shall be proven by contradiction, by showing that if $a^5b+3$ is a perfect cube then $ab^5+3$ cannot be.

Remark that perfect cubes are always congruent to $0$ , $1$ , or $-1$ modulo $9$ . Therefore, if $a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}$ , then $a^5b\equiv 5,6,\text{ or }7\pmod{9}$ .

If $a^5b\equiv 6\pmod 9$ , then note that $3|b$ . (This is because if $3|a$ then $a^5b\equiv 0\pmod 9$ .) Therefore $ab^5\equiv 0\pmod 9$ and $ab^5+3\equiv 3\pmod 9$ , contradiction.

Otherwise, either $a^5b\equiv 5\pmod 9$ or $a^5b\equiv 7\pmod 9$ . Note that since $a^6b^6$ is a perfect sixth power, and since neither $a$ nor $b$ contains a factor of $3$ , $a^6b^6\equiv 1\pmod 9$ . If $a^5b\equiv 5\pmod 9$ , then $a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.$ Similarly, if $a^5b\equiv 7\pmod 9$ , then $a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.$ Therefore $ab^5+3\equiv 5,7\pmod 9$ , contradiction.

Therefore no such integers exist.

Page

Toolbox

Search

2013 USAJMO Problems/Problem 1

Problem

Solution